package main

import "fmt"

/*
Given a 2D grid of size m x n and an integer k. You need to shift the grid k times.
In one shift operation:
Element at grid[i][j] moves to grid[i][j + 1].
Element at grid[i][n - 1] moves to grid[i + 1][0].
Element at grid[m - 1][n - 1] moves to grid[0][0].
Return the 2D grid after applying shift operation k times.

Example 1:
Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1
Output: [[9,1,2],[3,4,5],[6,7,8]]

Example 2:
Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]

Example 3:
Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9
Output: [[1,2,3],[4,5,6],[7,8,9]]

Constraints:
m == grid.length
n == grid[i].length
1 <= m <= 50
1 <= n <= 50
-1000 <= grid[i][j] <= 1000
0 <= k <= 100
*/

func main() {
	fmt.Println(shiftGrid([][]int{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 1))
	fmt.Println(shiftGrid([][]int{{3, 8, 1, 9}, {19, 7, 2, 5}, {4, 6, 11, 10}, {12, 0, 21, 13}}, 4))
}

func shiftGrid(grid [][]int, k int) [][]int {
	ret := make([][]int, len(grid))
	for i := 0; i < len(ret); i++ {
		ret[i] = make([]int, len(grid[0]))
	}

	for i := 0; i < len(grid); i++ {
		for j := 0; j < len(grid[0]); j++ {
			ret[((k+j)/len(grid[0])+i)%len(grid)][(k+j)%len(grid[0])] = grid[i][j]
		}
	}

	return ret
}
